Nepal Engineering Council · Mechanical Engineering · Chapter 7
Repair and Maintenance of Engineering System
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176 questions in 6 syllabus topics.
7.1 Need for maintenance
28 questions · AMeE0701
1. Maintenance of an engineering system is best defined as:
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Think about both keeping and restoring function.
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Answer: C. all actions needed to retain an item in, or restore it to, a state in which it can perform its required function
Standard definitions (e.g. BS EN 13306) describe maintenance as the combination of technical and administrative actions that keep an item in, or return it to, a state where it can do its required function.
2. Which of the following is the primary reason why every machine needs maintenance?
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What happens to rubbing surfaces and loaded parts over time?
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Answer: D. Progressive deterioration of components through wear, corrosion, fatigue and ageing
Components in service are subjected to friction, corrosion, cyclic loading and environmental attack, so their condition deteriorates with time and must be monitored and restored.
3. Which of the following is NOT a benefit of a good maintenance plan?
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Look for the outcome that is undesirable.
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Answer: B. Higher frequency of unplanned breakdowns
A good plan reduces unplanned breakdowns; an increase in them would indicate poor maintenance.
4. Wear of mating surfaces in a machine mainly results from:
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Consider what a lubricant film prevents.
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Answer: D. friction between surfaces in relative motion with inadequate lubrication
Relative motion under load with insufficient lubricant film causes abrasion and adhesion, removing material from the surfaces.
5. A machine is scheduled to work 160 h per month and was down for 12 h for repairs. Its availability for the month is:
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Availability = uptime / scheduled time.
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Answer: C. 92.5 %
Availability = (160 − 12)/160 = 148/160 = 0.925 = 92.5 %.
6. The difference between maintenance and repair is best stated as:
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One is ongoing and preventive, the other follows a fault.
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Answer: B. maintenance is a continuing activity to prevent failures, while repair restores a failed item to working condition
Maintenance covers planned and ongoing care to avoid failure, whereas repair is the corrective action after a fault has occurred.
7. A machine has an MTBF of 400 h and an MTTR of 25 h. Its inherent availability is nearly:
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Denominator is the full cycle: up time plus repair time.
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Answer: A. 94.1 %
A = MTBF/(MTBF + MTTR) = 400/425 = 0.941 = 94.1 %.
8. In the maintenance cost-versus-level curve, the total cost (maintenance cost + breakdown/downtime cost) is minimum at:
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One cost curve rises and the other falls.
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Answer: B. an intermediate (optimum) level of maintenance effort
Maintenance cost rises with effort while breakdown losses fall; their sum has a minimum at an optimum level.
9. Which of the following is a safety-related reason for maintaining pressure vessels and lifting equipment?
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Consider consequences of sudden failure of a pressurised or lifted load.
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Answer: A. Undetected defects can lead to explosion, collapse or injury, and inspection is often a legal requirement
Corrosion, cracks and fatigue in such equipment can cause catastrophic failure, so periodic inspection and upkeep protect people and satisfy regulations.
10. A plant has 18 breakdowns a year costing Rs 12,000 each. A preventive programme costing Rs 90,000 per year cuts breakdowns to 6 per year. The net annual saving is:
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Subtract the cost of the programme from the gross saving.
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Answer: B. Rs 54,000
Breakdowns avoided = 18 − 6 = 12, worth 12 × 12,000 = Rs 1,44,000; less programme cost Rs 90,000 gives Rs 54,000.
11. Which statement about maintenance and product quality is correct?
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Think about a worn spindle or slide on a machine tool.
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Answer: A. Well-maintained machines hold tolerances better, so scrap and rework are reduced
Wear and misalignment of machine tools lead to dimensional drift and defects; maintenance keeps the process capable.
12. A machine stops for 6 h each month. The lost contribution is Rs 4,500 per hour. The yearly cost of this downtime is:
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Convert monthly hours to yearly hours first.
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Answer: A. Rs 3,24,000
6 h × 12 months = 72 h; 72 × 4,500 = Rs 3,24,000.
13. Which of the following is an example of a hidden (indirect) cost of equipment breakdown?
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Direct costs appear on the repair job card.
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Answer: D. Loss of customer goodwill due to late delivery
Direct costs are labour and material of the repair; late deliveries and lost customers are indirect consequences that are harder to measure.
14. A pump failed 6 times in 3000 operating hours. Assuming a constant failure rate, its MTBF is:
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MTBF is operating time divided by number of failures.
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Answer: D. 500 h
λ = 6/3000 = 0.002 per hour; MTBF = 1/λ = 500 h.
15. Which of the following is a reason for maintenance arising from environmental factors?
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Which choice describes a harmful environment?
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Answer: A. Dust, humidity and chemical fumes accelerate corrosion and contamination of components
Contaminants, moisture and corrosive atmospheres damage surfaces and lubricants, increasing the need for cleaning, lubrication and protection.
16. Failures of a machine in 2000 h of operation fall after improved maintenance because MTBF increases from 200 h to 500 h. The number of failures avoided is:
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Failures = time / MTBF.
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Answer: D. 6
Expected failures before = 2000/200 = 10; after = 2000/500 = 4; avoided = 6.
17. An item that is kept in service beyond its economic life without renewal usually shows:
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Think of the wear-out region of the bathtub curve.
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Answer: C. rising maintenance cost and increasing frequency of failures
As equipment ages (wear-out phase), failures and repair costs rise, making replacement more economical.
18. Which of the following best describes an 'overhaul'?
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It is far more extensive than routine servicing.
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Answer: C. A comprehensive inspection and restoration of an item to an acceptable condition, usually by dismantling
An overhaul is an extensive, planned examination and restoration of equipment, often involving partial or full dismantling.
19. A major benefit of a good maintenance plan in terms of energy use is that:
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Consider friction and leakage losses.
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Answer: B. well-lubricated, aligned and clean machines consume less energy for the same output
Friction, leakage, fouling and misalignment waste energy; correcting them reduces consumption.
20. Which of the following is the main objective of a maintenance department?
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Think availability, cost and safety together.
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Answer: B. To maximise availability of plant at minimum total cost while meeting safety requirements
Maintenance supports production by keeping plant available, reliable and safe at an economical cost.
21. The ability of an item to be restored to its operating state within a given time using prescribed procedures is called:
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It concerns how easily repair can be done.
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Answer: A. maintainability
Maintainability is the probability that a failed item is restored within a specified time; reliability concerns failure-free operation.
22. A system with availability 0.90 is required to work all year (8760 h). The expected downtime per year is about:
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Downtime fraction is 1 minus availability.
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Answer: D. 876 h
Downtime = (1 − 0.90) × 8760 = 876 h.
23. Keeping records of failures, repairs and costs for each machine (equipment history) helps mainly to:
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Data supports decisions.
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Answer: B. decide on repair/replace policy and improve the maintenance schedule
History data reveals recurring faults and cost trends that support planning, spare holding and replacement decisions.
24. A company spends Rs 1.8 million per year on maintenance and the annual output is valued at Rs 45 million. Maintenance cost as a percentage of output value is:
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Divide cost by output value, then multiply by 100.
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Answer: D. 4 %
(1.8/45) × 100 = 4 %.
25. Failure to maintain a machine's guards, interlocks and brakes mainly increases the risk of:
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Think about the purpose of a guard.
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Answer: C. accidents and injuries to operators
Defective safety devices expose people to hazards, which is a key legal and ethical reason for maintenance.
26. Which of the following is a benefit of maintaining spare parts and tools in an organised store as part of a maintenance plan?
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Which part of the repair time does stock affect?
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Answer: A. Reduced repair time and hence reduced downtime
Availability of the right spares shortens waiting time during repair, reducing MTTR.
27. Deferring all maintenance to cut short-term cost typically results in:
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Compare short-term and long-term cost.
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Answer: C. higher long-term cost because of failures, lost production and shortened equipment life
Savings from skipped maintenance are usually outweighed by later breakdown and replacement costs.
28. Fatigue is a reason for maintenance because it:
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Repeated stress cycles are involved.
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Answer: C. causes cracks to initiate and grow under repeated loading, eventually leading to sudden fracture
Cyclic stresses well below the ultimate strength can produce progressive cracking, so critical parts are inspected and replaced.
7.2 Safety precautions and protective coating
29 questions · AMeE0702
29. The three elements of the fire triangle are:
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Starving, cooling and smothering correspond to the three sides.
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Answer: D. fuel, heat (ignition source) and oxygen
Combustion needs fuel, an ignition source and an oxidiser (oxygen); removing any one extinguishes the fire.
30. Under the common (international) classification, Class B fires involve:
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Class A is wood and paper; B is the next group.
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Answer: A. flammable liquids such as petrol, oil and paints
Class A is ordinary solids, Class B flammable liquids, Class C gases (in some systems), Class D metals; electrical fires are treated separately.
31. Which extinguishing agent is most suitable for a fire in live electrical switchgear?
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The agent must not conduct electricity.
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Answer: C. Carbon dioxide (CO2) extinguisher
CO2 is non-conductive and leaves no residue; water and foam conduct electricity and can cause shock.
32. Fires involving combustible metals such as magnesium or titanium are classified as:
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Letters run from ordinary solids to metals.
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Answer: A. Class D
Class D fires require special dry powder agents; water can react violently with burning metals.
33. The first action on discovering a fire on electrical equipment that has been energised should be to:
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Remove the energy source and warn others.
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Answer: B. raise the alarm and, if safe, isolate the electrical supply
Isolating the supply removes the electrical ignition source and the shock hazard while the alarm brings help.
34. A person with a body resistance of 1500 Ω touches a 230 V live conductor. The current through the body is about:
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Use Ohm's law and watch the units.
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Answer: B. 153 mA
I = V/R = 230/1500 = 0.153 A = 153 mA, well within the range that can cause ventricular fibrillation.
35. A residual current device (RCD) protects against electric shock by:
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It compares the current going out with the current coming back.
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Answer: B. disconnecting the supply when the line and neutral currents differ by more than a set leakage value
An RCD senses imbalance between phase and neutral currents (leakage to earth) and trips quickly, commonly at 30 mA for personal protection.
36. Lock-out/tag-out (LOTO) procedures are used during maintenance in order to:
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The lock stays with the person doing the work.
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Answer: D. prevent accidental energising or start-up of machinery while work is in progress
Isolating energy sources and locking them with tags ensures no one can restart the equipment while someone is working on it.
37. Which of the following is the main purpose of earthing metallic frames of electrical machines?
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Think about what happens if a live conductor touches the frame.
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Answer: B. To ensure that a fault current flows to earth, operating the protection and limiting touch voltage
If insulation fails, the frame stays near earth potential and the fault current trips the protective device.
38. A guard on a machine tool that is interlocked with the drive is designed so that:
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The name suggests a linkage between guard and power.
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Answer: A. the machine cannot run while the guard is open or removed
Interlocking links the guard to the control circuit, stopping the drive (or preventing start) when the guard is opened.
39. Which of the following is an unsafe practice when operating a lathe?
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Rotating parts catch loose items.
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Answer: A. Wearing loose clothing, gloves or ties near the rotating chuck
Loose items can be caught by rotating parts and pull the operator into the machine.
40. A grinding wheel of diameter 250 mm runs at 3000 rpm. Its peripheral speed is about:
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Convert diameter to metres and rpm to rev/s.
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Answer: A. 39.3 m/s
v = πDN/60 = π × 0.25 × 3000/60 = 39.3 m/s; this must not exceed the wheel's marked maximum safe speed.
41. Corrosion that occurs when two dissimilar metals are in electrical contact in an electrolyte is called:
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A battery-like cell is formed.
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Answer: A. galvanic corrosion
The more anodic metal corrodes preferentially when coupled with a more noble metal in a conducting medium.
42. Localised corrosion that produces small cavities or holes in a metal surface is termed:
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Small holes.
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Answer: B. pitting corrosion
Pitting is an intense localised attack, common in stainless steels exposed to chlorides.
43. Crevice corrosion is most likely to occur:
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Think of stagnant solution in a narrow space.
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Answer: D. in narrow gaps where the electrolyte is stagnant, such as under gaskets and washers
Oxygen depletion inside a crevice sets up a differential aeration cell, attacking the metal within the gap.
44. Sensitisation of austenitic stainless steel, which leads to intergranular corrosion, is caused by:
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Welding heat affected zones are often attacked.
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Answer: C. precipitation of chromium carbides at grain boundaries when heated in the range of about 500–800 °C
Chromium carbides at grain boundaries deplete adjacent regions of chromium, making them vulnerable; Ti or Nb additions (stabilisation) prevent this.
45. Stress corrosion cracking requires the simultaneous presence of:
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Three conditions must coincide.
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Answer: D. a tensile stress, a susceptible material and a specific corrosive environment
SCC arises from the combined action of sustained tensile stress (applied or residual) and a particular environment, e.g. brass in ammonia.
46. Dezincification is a form of selective leaching that occurs in:
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The name contains the element lost.
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Answer: C. brass, where zinc is preferentially removed leaving a porous copper-rich structure
In brasses with high zinc content, zinc dissolves leaving spongy copper that is weak and permeable.
47. Which of the following is a corrosion inhibitor action?
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Small additions to the fluid.
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Answer: D. Forming a thin protective film on the metal surface that slows the anodic or cathodic reaction
Inhibitors are added in small quantities to the environment and adsorb on the surface or form films that reduce the corrosion rate.
48. Steel loses 3.925 g from an exposed area of 100 cm² uniformly over one year (density 7.85 g/cm³). The corrosion rate in mm/year is:
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Thickness = mass / (density × area); convert cm to mm.
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Answer: C. 0.05
Thickness lost = m/(ρA) = 3.925/(7.85 × 100) = 0.005 cm = 0.05 mm per year.
49. Iron dissolves (Fe → Fe²⁺ + 2e⁻) at a steady corrosion current of 0.5 A for 10 hours. Using M = 55.85 g/mol and F = 96485 C/mol, the mass of iron lost is about:
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Faraday's law with n = 2.
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Answer: C. 5.2 g
m = M I t/(nF) = 55.85 × 0.5 × 36000/(2 × 96485) = 5.21 g.
50. The zinc coating on a galvanised sheet has a mass of 600 g/m² (zinc density 7140 kg/m³). The coating thickness is about:
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Thickness = mass per unit area ÷ density.
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Answer: B. 84 µm
t = mass per area/density = 0.6 kg/m² ÷ 7140 kg/m³ = 8.4 × 10⁻⁵ m = 84 µm.
51. Galvanised (zinc-coated) steel continues to protect steel even where the coating is scratched because:
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Compare the positions of Zn and Fe in the galvanic series.
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Answer: D. zinc is more anodic than steel and corrodes sacrificially
Zinc is higher in the galvanic series (more active) than iron, so it becomes the anode and protects the exposed steel cathodically.
52. A paint has 60 % volume solids. For a required dry film thickness of 75 µm, the theoretical spreading rate is:
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Coverage = 10 × VS% / DFT in µm.
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Answer: C. 8 m²/L
Theoretical coverage (m²/L) = volume solids (%) × 10/DFT (µm) = 60 × 10/75 = 8 m²/L.
53. Which of the following materials is used as a sacrificial anode for the cathodic protection of buried steel pipelines?
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It must be more active than iron.
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Answer: D. Magnesium
Magnesium, zinc and aluminium are more active than steel and supply protective current by dissolving.
54. In impressed current cathodic protection (ICCP), the structure to be protected is connected to:
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The protected structure must be the cathode.
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Answer: A. the negative terminal of a DC source, with inert anodes connected to the positive terminal
A rectifier forces current from inert anodes (e.g. graphite) through the soil to the structure, making it the cathode.
55. A zinc sacrificial anode of 10 kg with a capacity of 780 A·h/kg and utilisation factor 0.85 supplies a steady 0.5 A. Its life is about:
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Life = usable charge ÷ current; convert hours to years.
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Answer: A. 1.5 years
Life = 10 × 780 × 0.85/0.5 = 13,260 h = 13,260/8760 ≈ 1.5 years.
56. Phosphate treatment of steel before painting is mainly done to:
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Pre-treatment improves bonding of coating.
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Answer: B. form a thin crystalline layer that improves paint adhesion and gives corrosion resistance
Phosphating converts the surface to an adherent phosphate layer that anchors paint and slows underfilm corrosion.
57. The standard criterion for adequate cathodic protection of steel in soil is a structure-to-electrolyte potential, measured against a Cu/CuSO4 reference electrode, of at least:
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It is a negative value.
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Answer: C. −0.85 V (more negative)
A potential of −0.85 V vs Cu/CuSO4 or more negative is the widely used criterion that steel is cathodically protected.
7.3 Maintenance strategy
29 questions · AMeE0703
58. Which of the following is NOT normally a factor in choosing a maintenance strategy for an item of equipment?
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Look for the irrelevant factor.
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Answer: A. The colour of the machine paint
Strategy is chosen on criticality, failure pattern, cost, safety consequences and resources; appearance has no bearing.
59. Preventive maintenance (PM) is best described as maintenance:
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Think of the word 'prevent' and a plan.
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Answer: C. carried out at predetermined intervals or according to prescribed criteria to reduce the probability of failure
PM consists of planned inspection, servicing and replacement actions intended to prevent failures before they happen.
60. Breakdown (run-to-failure) maintenance is most appropriate for:
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Which item can fail without any harm?
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Answer: C. non-critical, low-cost items whose failure has no safety consequence, such as a lamp
Where failure is inexpensive, harmless and easily remedied, waiting for failure is the cheapest option.
61. A major drawback of breakdown maintenance is that:
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Think of what 'unexpected' implies.
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Answer: B. failure occurs unexpectedly, causing production loss and sometimes secondary damage to other parts
Unplanned failures stop production at inconvenient times and may damage associated components, increasing repair costs.
62. Scheduled maintenance is carried out:
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Think of a calendar or hour-meter.
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Answer: D. at fixed calendar or operating-hour intervals irrespective of the actual condition of the item
Scheduled maintenance is time- or usage-based, e.g. oil change every 500 h, without checking the real condition first.
63. A lubrication task is scheduled every 250 operating hours. A machine runs 400 operating hours per month. How many times is the task done in a year (approximately)?
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Divide the yearly operating hours by the interval.
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Answer: B. 19
Operating hours per year = 400 × 12 = 4800; 4800/250 = 19.2, i.e. about 19 times.
64. A disadvantage of purely time-based preventive maintenance is that it:
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It ignores the true state of the item.
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Answer: A. may replace components that are still in good condition and does not prevent random failures
Fixed-interval replacement wastes remaining life and is of little help for failures that occur at random.
65. Total Productive Maintenance (TPM) is a philosophy that:
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The word 'total' indicates everyone participates.
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Answer: B. involves all employees, including operators, in maintaining equipment to maximise its overall effectiveness
TPM, developed in Japan, is a company-wide, team-based approach in which operators share responsibility for care of their machines.
66. In TPM, 'autonomous maintenance' (Jishu Hozen) means:
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Who does the work?
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Answer: B. operators perform basic cleaning, lubrication and inspection of their own equipment
Operators take ownership of routine care tasks, detecting abnormalities early and freeing technicians for skilled work.
67. The three 'zero' goals commonly associated with TPM are zero:
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Think equipment, product and people.
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Answer: B. breakdowns, defects and accidents
TPM aims at zero unplanned breakdowns, zero defects (quality) and zero accidents.
68. Overall Equipment Effectiveness (OEE) is calculated as the product of:
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Three percentages are multiplied.
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Answer: A. availability, performance rate and quality rate
OEE = A × P × Q, measuring how much of the planned time is spent producing good parts at the ideal rate.
69. A machine has availability 90 %, performance rate 95 % and quality rate 98 %. Its OEE is:
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Multiply the three factors; do not average them.
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Answer: C. 83.8 %
OEE = 0.90 × 0.95 × 0.98 = 0.838 = 83.8 %.
70. In an 8-hour (480 min) shift, a machine is down for 48 min, has an ideal cycle time of 0.4 min, produces 1000 parts and 980 are good. The OEE is about:
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Performance uses ideal time for all parts divided by run time.
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Answer: D. 81.7 %
A = 432/480 = 0.90; P = (0.4 × 1000)/432 = 0.926; Q = 980/1000 = 0.98; OEE = 0.90 × 0.926 × 0.98 = 0.817.
71. Which of the following is one of the six big losses addressed by TPM?
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Think of losses that reduce OEE.
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Answer: C. Reduced speed and minor stoppages
The six big losses are breakdowns, set-up/adjustments, idling and minor stops, reduced speed, process defects and start-up/yield losses.
72. In the TPM framework, the 'quality maintenance' pillar aims to:
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Defects are traced to equipment conditions.
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Answer: B. keep equipment conditions under control so that defect-free products are made
Quality maintenance links machine conditions to product quality, preventing defects by controlling the equipment parameters that affect quality.
73. Condition monitoring differs from time-based maintenance mainly because it:
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The word 'condition' is key.
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Answer: D. uses the measured condition of the equipment to decide when maintenance is needed
Condition-based maintenance acts on measured indicators (vibration, temperature, oil debris), avoiding unnecessary work and catching deterioration early.
74. Vibration analysis is most useful for detecting:
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The machine must have moving parts.
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Answer: D. unbalance, misalignment and bearing defects in rotating machinery
Characteristic frequencies in the vibration spectrum indicate faults such as unbalance (1× rpm), misalignment (2× rpm) and bearing defects.
75. A shaft rotates at 1500 rpm. Vibration due to unbalance appears at a frequency of:
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Convert rpm to revolutions per second.
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Answer: C. 25 Hz
Unbalance vibrates at once-per-revolution: 1500/60 = 25 Hz.
76. A gearbox pinion with 40 teeth rotates at 1500 rpm. The gear-mesh frequency is:
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Multiply teeth count by revolutions per second.
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Answer: B. 1000 Hz
Mesh frequency = teeth × shaft speed = 40 × (1500/60) = 1000 Hz.
77. Infrared thermography is used in condition monitoring mainly to:
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It senses radiated heat.
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Answer: D. detect abnormal heating, e.g. loose electrical connections or overheated bearings
Faults generate excess heat; thermal images show hot spots before failure occurs.
78. Analysis of wear particles (ferrography) in used lubricating oil indicates:
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Oil carries evidence of what is wearing inside.
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Answer: A. the type and rate of wear occurring inside the machine
Quantity, size and shape of particles reveal which components are wearing and how severely.
79. Without PM a machine has 5 failures a year costing Rs 20,000 each. With a PM programme costing Rs 40,000 per year there are 2 failures per year. The annual saving from PM is:
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Compare total yearly cost in both cases.
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Answer: D. Rs 20,000
Without PM: 5 × 20,000 = 1,00,000. With PM: 40,000 + 2 × 20,000 = 80,000. Saving = Rs 20,000.
80. A maintenance section has 120 jobs per month averaging 6 hours each. A technician provides 150 productive hours per month. The minimum number of technicians needed is:
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Round up since technicians cannot be fractional.
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Answer: C. 5
Workload = 120 × 6 = 720 h; 720/150 = 4.8, rounded up to 5 technicians.
81. A component has an exponential life with MTBF 1000 h. The time to which PM should be scheduled to keep reliability at 90 % is about:
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Use R = exp(−t/MTBF) and take natural log.
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Answer: C. 105 h
R = e^(−t/MTBF) = 0.90 → t = −1000 ln 0.90 = 105 h.
82. A condition-monitoring system costs Rs 2,00,000 and is expected to prevent 4 failures a year, each costing Rs 75,000. The payback period is:
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Divide investment by annual saving.
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Answer: D. 8 months
Annual saving = 4 × 75,000 = Rs 3,00,000; payback = 2,00,000/3,00,000 = 0.667 year = 8 months.
83. Total planned quality maintenance and the quality-maintenance approach of TPM are most concerned with:
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Planned upkeep and product quality are combined.
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Answer: A. linking the planned upkeep of equipment to the quality of the products it makes
These approaches plan maintenance so that equipment conditions stay within limits that guarantee product quality.
84. Predictive maintenance is best described as:
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It forecasts failure from trends.
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Answer: A. maintenance initiated by the predicted deterioration of an item from analysis of measured parameters
Trend data (vibration, temperature, oil analysis) is used to forecast the remaining useful life and plan work just in time.
85. For equipment whose failures occur at random with a constant failure rate, fixed-interval replacement is generally:
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Constant failure rate means no wear-out.
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Answer: A. ineffective, since the item is no more likely to fail after a long time than after a short time
With a constant hazard rate, age gives no warning of failure, so time-based replacement does not lower the failure probability.
86. Opportunity maintenance means:
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Use the stoppage that already exists.
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Answer: A. carrying out extra maintenance tasks when the plant is stopped for another reason
When equipment is shut for another reason, otherwise inconvenient tasks are done to save additional downtime.
7.4 System safety and reliability
30 questions · AMeE0704
87. The bathtub curve of failure rate versus time for equipment consists of which three periods, in order?
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Early, middle, late life.
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Answer: C. infant mortality, useful life (constant failure rate), wear-out
The failure rate is high and decreasing early on, roughly constant in mid-life, then rising as components wear out.
88. Early-life (infant mortality) failures are mainly caused by:
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Think of quality problems at the start of life.
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Answer: B. manufacturing defects, poor assembly or installation errors
Weak components and assembly or installation mistakes fail soon after commissioning; burn-in testing weeds them out.
89. During the useful-life period of the bathtub curve, failures are mostly:
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The flat part of the bathtub.
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Answer: A. random, with an approximately constant failure rate
In mid-life, failures occur at random from chance overloads and are modelled by the exponential distribution.
90. For a Weibull distribution of failure times, a shape parameter β < 1 indicates:
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β = 1 is the exponential case.
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Answer: A. a decreasing failure rate, typical of infant mortality
β < 1 gives decreasing hazard, β = 1 constant (exponential), and β > 1 increasing hazard.
91. A component has a constant failure rate of 0.001 per hour. Its reliability over 200 h is:
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Use the exponential reliability function.
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Answer: A. 0.819
R = e^(−λt) = e^(−0.2) = 0.819.
92. For a constant failure rate of 0.0005 per hour, the MTBF is:
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MTBF is the reciprocal of the failure rate.
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Answer: C. 2000 h
MTBF = 1/λ = 1/0.0005 = 2000 h.
93. The probability that a component with exponential life survives up to a time equal to its MTBF is:
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Substitute t = MTBF in e^(−t/MTBF).
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Answer: B. 0.368
R(MTBF) = e^(−1) = 0.368.
94. Three independent components of reliabilities 0.95, 0.90 and 0.98 are connected in series. The system reliability is:
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Series systems multiply reliabilities.
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Answer: A. 0.838
R = 0.95 × 0.90 × 0.98 = 0.8379.
95. Two independent units of reliabilities 0.90 and 0.80 are connected in parallel (either one sufficient). The system reliability is:
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Parallel: one minus product of unreliabilities.
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Answer: D. 0.98
R = 1 − (1 − 0.90)(1 − 0.80) = 1 − 0.02 = 0.98.
96. A unit A (R = 0.90) is in series with two parallel units B and C (R = 0.80 each). The reliability of the system is:
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Reduce the parallel block first.
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Answer: C. 0.864
Parallel B–C: 1 − 0.2² = 0.96; series with A: 0.90 × 0.96 = 0.864.
97. Two identical units, each with failure rate 0.001 per hour, operate in active parallel redundancy (system fails when both fail; no repair). The system MTTF is:
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MTTF = (1 + 1/2)/λ.
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Answer: D. 1500 h
For two identical parallel units, MTTF = 1/λ + 1/(2λ) = 1000 + 500 = 1500 h.
98. Three components in series have constant failure rates of 0.0002 per hour each. The MTBF of the system is about:
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Add the failure rates for a series system.
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Answer: A. 1667 h
λsystem = 3 × 0.0002 = 0.0006 per hour; MTBF = 1/0.0006 = 1667 h.
99. In a test, 10 identical units run for 500 h each and 5 failures occur. Assuming a constant failure rate, the estimated failure rate is:
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Failures divided by total accumulated hours.
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Answer: D. 0.001 per hour
Total operating time = 10 × 500 = 5000 unit-hours; λ = 5/5000 = 0.001 per hour.
100. A Weibull model has scale η = 1000 h and shape β = 2. The reliability at 500 h is:
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Square the ratio t/η.
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Answer: B. 0.779
R = exp[−(t/η)^β] = exp[−(0.5)²] = e^(−0.25) = 0.779.
101. Mean time to failure (MTTF) is usually used for non-repairable items, while for repairable items the mean time between failures is related by:
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Cycle = up time + repair time.
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Answer: B. MTBF = MTTF + MTTR
One full operating cycle of a repairable item includes the up time (MTTF) plus the mean repair time (MTTR).
102. A repairable machine has MTTF = 900 h and MTTR = 100 h. Its steady-state availability is:
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Up time over total cycle.
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Answer: B. 0.90
A = MTTF/(MTTF + MTTR) = 900/1000 = 0.90.
103. Which one of the following is a failure mode of a mechanical component?
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A mode describes how it fails.
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Answer: A. Fatigue fracture
A failure mode is the manner in which an item fails, e.g. fatigue, wear, creep, buckling, corrosion or brittle fracture; the others are performance measures.
104. Gradual, time-dependent plastic deformation of a metal under constant load at high temperature is the failure mode called:
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Think of turbine blades at high temperature.
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Answer: D. creep
Creep occurs at elevated temperature (above about 0.4 Tm) under sustained stress, e.g. in turbine blades and boiler tubes.
105. Failure Mode and Effects Analysis (FMEA) is a:
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It starts from component level.
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Answer: A. systematic bottom-up method that identifies failure modes of components and their effects on the system
FMEA examines each component's possible failure modes and evaluates the effects, causes and detection.
106. In an FMEA, a failure mode has severity 8, occurrence 4 and detection 5. The Risk Priority Number (RPN) is:
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Multiply the three ratings.
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Answer: C. 160
RPN = S × O × D = 8 × 4 × 5 = 160.
107. Fault Tree Analysis (FTA) is characterised by being:
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It works backwards from a failure.
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Answer: B. a top-down deductive method that starts with an undesired top event and traces its causes
FTA starts from the system failure (top event) and uses logic gates to find the combinations of events that cause it.
108. In a fault tree, two independent basic events with probabilities 0.1 and 0.2 are connected to an AND gate. The output event probability is:
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AND means both must occur.
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Answer: D. 0.02
AND gate: P = 0.1 × 0.2 = 0.02.
109. Three independent basic events each with probability 0.01 feed an OR gate. The probability of the output event is about:
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OR gate: one minus the product of non-occurrence probabilities.
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Answer: D. 0.0297
P = 1 − (1 − 0.01)³ = 1 − 0.970299 = 0.0297.
110. Event Tree Analysis (ETA) differs from FTA in that ETA:
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Forward logic from an initiating event.
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Answer: C. starts with an initiating event and follows the success/failure of safety systems forward to the possible outcomes
ETA is an inductive forward-logic technique that branches through safeguards to determine the outcome frequencies.
111. In an event tree, an initiating event occurs 0.1 times per year and the single safeguard fails with probability 0.05. The frequency of the 'safeguard fails' outcome is:
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Multiply initiating frequency by branch probability.
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Answer: B. 0.005 per year
Frequency = 0.1 × 0.05 = 0.005 per year.
112. HAZOP (Hazard and Operability study) examines a process by applying:
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Look for the structured use of keywords.
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Answer: C. guide words such as 'no', 'more' and 'less' to process parameters to find deviations
A team applies guide words to parameters (flow, pressure, temperature) to identify deviations, their causes and consequences.
113. A fishbone (Ishikawa, cause-and-effect) diagram is used to:
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The 'head' of the fish holds the problem.
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Answer: A. identify and organise the possible causes of a specific problem or effect
The effect is placed at the head of the 'fish' and causes are grouped on the bones, commonly as man, machine, material, method, measurement and environment.
114. A Reliability Block Diagram (RBD) represents:
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It is about success paths, not geometry.
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Answer: B. the logical connection (series/parallel) of components required for the system to function
In an RBD, success paths through blocks show which component failures cause system failure.
115. Which of the following lists correctly the usual categories of causes of equipment failure?
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More than one origin is possible.
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Answer: C. design faults, material defects, manufacturing/installation errors, operating misuse and inadequate maintenance
Failures commonly have several contributing causes spread across design, materials, manufacture, operation and maintenance.
116. In system safety, a 'hazard' is best defined as:
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Hazard is the source; risk includes likelihood.
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Answer: D. a condition or situation with the potential to cause harm, damage or loss
A hazard is the source of potential harm; risk combines its likelihood and severity.
7.5 Risk assessment
30 questions · AMeE0705
117. In risk assessment, risk is commonly expressed as:
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Two factors are multiplied.
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Answer: C. the product of the likelihood (probability) of an event and the magnitude of its consequence
Risk = probability (or frequency) × consequence (severity); it measures expected loss.
118. Which statement correctly distinguishes hazard from risk?
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One is a source, the other a measure.
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Answer: D. A hazard is a source of potential harm, while risk is the likelihood of that harm combined with its severity
A wet floor is a hazard; the risk depends on how likely a fall is and how serious the injury would be.
119. An event has an annual probability of 0.02 and would cause a loss of Rs 5,00,000. The annual expected loss (risk) is:
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Multiply probability by consequence.
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Answer: B. Rs 10,000
Risk = 0.02 × 5,00,000 = Rs 10,000 per year.
120. Hazard A: probability 0.05 per year, loss Rs 4,00,000. Hazard B: probability 0.01 per year, loss Rs 30,00,000. The total expected annual loss from both is:
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Compute each risk separately and add.
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Answer: B. Rs 50,000
A: 0.05 × 4,00,000 = 20,000; B: 0.01 × 30,00,000 = 30,000; total = Rs 50,000.
121. On a 5 × 5 risk matrix, a hazard has likelihood rating 4 and severity rating 5. Its risk score is:
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Multiply the two ratings.
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Answer: C. 20
Score = likelihood × severity = 4 × 5 = 20.
122. Which risk analysis approach ranks risks using descriptive categories such as low, medium and high based on judgement?
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No numbers are calculated.
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Answer: B. Qualitative risk analysis
Qualitative methods use descriptive scales and expert judgement; quantitative methods use numerical probabilities and consequences.
123. Which of the following is a quantitative risk analysis technique?
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It must yield numbers.
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Answer: A. Fault tree analysis with probability calculation
Fault trees can be quantified using the probabilities of basic events to compute the top-event frequency.
124. A 'What-if' analysis for safety is:
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It is a brainstorming method.
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Answer: C. a structured brainstorming of questions beginning 'What if…?' to identify hazards and consequences
The team asks 'what if' questions about deviations, failures and errors and records consequences and safeguards.
125. ALARP in risk management stands for:
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Reasonableness and practicality limit further reduction.
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Answer: D. as low as reasonably practicable
ALARP means risk must be reduced until the further cost or effort is grossly disproportionate to the benefit gained.
126. According to the hierarchy of controls, the MOST effective way of reducing a risk is to:
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Remove it altogether.
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Answer: D. eliminate the hazard
The order of effectiveness is elimination, substitution, engineering controls, administrative controls and PPE.
127. In the hierarchy of controls, personal protective equipment (PPE) is:
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It is at the bottom of the hierarchy.
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Answer: D. the last line of defence and the least effective control
PPE only protects the wearer and depends on correct use, so it is used when higher-level controls cannot reduce risk enough.
128. Replacing a toxic solvent by a less harmful one is an example of risk reduction by:
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It is the second step in the hierarchy.
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Answer: A. substitution
Substitution replaces the hazard with a less hazardous alternative.
129. Fitting a fixed guard around a rotating shaft is an example of an:
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It physically separates people from the hazard.
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Answer: B. engineering control
Physical guards isolate people from the hazard without relying on behaviour.
130. Taking out an insurance policy against fire loss is an example of risk:
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Someone else bears the financial loss.
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Answer: C. transfer
Insurance transfers the financial consequences of the risk to another party; the hazard itself still exists.
131. The risk that remains after control measures have been applied is called:
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It is what is 'left over'.
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Answer: D. residual risk
Residual risk is what is left after controls; it must be tolerable (ALARP).
132. An unmitigated risk score of 100 units is reduced by 85 % by control measures. The residual risk is:
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Subtract the reduction from the original.
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Answer: A. 15 units
Residual = 100 × (1 − 0.85) = 15 units.
133. A control measure lowers the probability of an event from 0.04 to 0.01 per year. The loss if the event occurs is Rs 10,00,000, and the control costs Rs 20,000 per year. The net annual benefit is:
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Compare the risk reduction with the cost.
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Answer: B. Rs 10,000
Risk reduction = (0.04 − 0.01) × 10,00,000 = Rs 30,000; net benefit = 30,000 − 20,000 = Rs 10,000.
134. Two independent hazards have probabilities of 0.1 and 0.2 in the same year. The probability that at least one occurs is:
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One minus the probability that neither occurs.
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Answer: B. 0.28
P = 1 − (1 − 0.1)(1 − 0.2) = 1 − 0.72 = 0.28.
135. In a layer of protection analysis, an initiating event occurs 0.1 times per year and two independent protection layers each have a probability of failure on demand of 0.1. The mitigated event frequency is:
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Multiply the initiating frequency by each layer's PFD.
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Answer: A. 0.001 per year
Frequency = 0.1 × 0.1 × 0.1 = 0.001 per year.
136. The five steps of a typical workplace risk assessment (e.g. as recommended by HSE) start with:
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You cannot assess what you have not identified.
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Answer: A. identifying the hazards
The steps are: identify hazards; decide who might be harmed and how; evaluate risks and decide on controls; record the findings; review and update.
137. An industry records 6 lost-time injuries in 1.2 million man-hours worked. The frequency rate (injuries per million man-hours) is:
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Scale the injuries to one million hours.
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Answer: D. 5
Frequency rate = 6 × 10⁶/1.2 × 10⁶ = 5.
138. A factory loses 240 working days to injuries in 800,000 man-hours. The severity rate (days lost per million man-hours) is:
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Scale days lost to one million hours.
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Answer: A. 300
Severity rate = 240 × 10⁶/800,000 = 300.
139. A workforce of 600 had 9 reportable injuries in a year. The incidence rate per 100 workers is:
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Divide by workers and multiply by 100.
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Answer: A. 1.5
Incidence rate = 9/600 × 100 = 1.5.
140. There were 2 fatalities among workers exposed for a total of 4 × 10⁷ hours. The fatal accident rate (FAR, per 10⁸ exposure hours) is:
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Scale the fatalities to 100 million hours.
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Answer: C. 5
FAR = 2 × 10⁸/(4 × 10⁷) = 5.
141. A 'near miss' in industrial safety is:
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Nothing was harmed, but it could have been.
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Answer: B. an unplanned event that did not result in injury or damage but had the potential to do so
Near-miss reporting provides warnings of latent hazards before they cause actual harm.
142. A 'permit-to-work' system in an industry is used mainly to:
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A document that authorises dangerous work.
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Answer: D. control hazardous work, such as hot work or confined space entry, through formal written authorisation and precautions
A permit ensures hazards are identified, isolations are in place and the work is coordinated before it starts.
143. Job Safety Analysis (JSA) is carried out by:
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Step-by-step review of a task.
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Answer: A. breaking a job into steps, identifying the hazards in each step and deciding on controls
JSA examines each task step to find hazards and write safe work procedures.
144. Heinrich's accident triangle (1 : 29 : 300) suggests that for each major injury there are about:
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The base of the triangle is the largest number.
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Answer: B. 29 minor injuries and 300 no-injury incidents
The ratio illustrates that serious injuries are the tip of a base of many minor injuries and unsafe incidents.
145. A Safety Data Sheet (SDS/MSDS) for a chemical gives information on:
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Information on safe use.
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Answer: C. hazards, safe handling, storage, first aid and emergency measures
The SDS is a standardised document supplying hazard and precautionary information about a substance.
146. The term 'acceptable (tolerable) risk' refers to:
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Zero risk is not practical.
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Answer: C. a level of risk that society or the organisation is willing to live with in view of the benefits and costs of further reduction
Zero risk is not attainable; the tolerable level is set through criteria such as ALARP.
7.6 Non-destructive techniques (NDT)
30 questions · AMeE0706
147. Non-destructive testing (NDT) is a group of techniques used to:
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The word 'non-destructive' tells you the item is not damaged.
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Answer: C. examine a component for flaws or properties without impairing its future usefulness
NDT evaluates integrity without damaging the part, so the inspected item can remain in service.
148. Ultrasonic testing uses sound waves of frequency typically in the range of:
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Much higher than the audible range.
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Answer: B. 0.5 to 25 MHz
Ultrasonic inspection uses high-frequency sound, well above the audible range, commonly 0.5–25 MHz.
149. The device that generates and detects ultrasonic waves in a probe is a:
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It converts between electrical and mechanical energy.
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Answer: C. piezoelectric crystal (transducer)
A piezoelectric element converts electrical pulses to mechanical vibrations and vice versa.
150. A couplant (such as gel or oil) is applied between an ultrasonic probe and the test surface in order to:
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Air is a poor transmitter of ultrasound.
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Answer: D. remove the air gap so that the sound energy can enter the specimen
Air has very high impedance mismatch with solids and reflects nearly all ultrasound, so a liquid couplant is required.
151. In pulse-echo testing of a steel plate (longitudinal wave velocity 5900 m/s), the back-wall echo arrives 10 µs after the pulse is sent. The plate thickness is:
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The measured time is for a round trip.
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Answer: B. 29.5 mm
The sound travels to the back wall and returns: t = vT/2 = 5900 × 10 × 10⁻⁶/2 = 0.0295 m = 29.5 mm.
152. In steel (v = 5900 m/s) an echo from a flaw is received 6 µs after the pulse. The depth of the flaw is:
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Halve the round-trip distance.
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Answer: D. 17.7 mm
Depth = vT/2 = 5900 × 6 × 10⁻⁶/2 = 17.7 mm.
153. The wavelength of a 2 MHz longitudinal ultrasonic wave in steel (v = 5900 m/s) is:
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λ = v / f.
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Answer: A. 2.95 mm
λ = v/f = 5900/(2 × 10⁶) = 2.95 × 10⁻³ m = 2.95 mm.
154. The acoustic impedance of aluminium (ρ = 2700 kg/m³, v = 6320 m/s) is about:
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Z is the product of density and velocity.
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Answer: C. 17.1 MRayl
Z = ρv = 2700 × 6320 = 1.71 × 10⁷ kg/m²s = 17.1 MRayl.
155. The acoustic impedance of steel is 46.3 MRayl and of water 1.48 MRayl. At a water–steel interface the fraction of incident energy reflected at normal incidence is about:
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Large impedance mismatch gives high reflection.
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Answer: A. 88 %
R = [(Z2 − Z1)/(Z2 + Z1)]² = (44.82/47.78)² = 0.88.
156. A 10 mm diameter, 2 MHz transducer is used on steel (v = 5900 m/s). The near-field length N = D²f/4v is about:
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Use SI units for all quantities.
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Answer: D. 8.5 mm
N = (0.01)² × 2 × 10⁶/(4 × 5900) = 200/23600 = 8.5 × 10⁻³ m = 8.5 mm.
157. Increasing the frequency of an ultrasonic probe generally:
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There is a trade-off between resolution and penetration.
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Answer: D. improves the resolution and sensitivity to small flaws but increases attenuation
Shorter wavelength resolves smaller defects, but scattering and absorption increase, limiting penetration depth.
158. Angle (shear wave) probes are mainly used in ultrasonic testing of:
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The weld cap prevents placing a straight probe on the weld itself.
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Answer: C. welds, to detect planar defects such as cracks and lack of fusion
Beams introduced at an angle through the weld region can intercept flaws that lie perpendicular to the beam.
159. The sequence of steps in liquid (dye) penetrant testing is:
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The developer comes after the excess penetrant is removed.
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Answer: A. clean, apply penetrant, dwell, remove excess, apply developer, inspect
The surface is cleaned, penetrant applied and left to dwell, excess is removed, then developer draws the trapped penetrant out for inspection.
160. The physical principle that draws the penetrant into a surface-breaking crack in dye penetrant testing is:
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Think of liquid rising in a narrow tube.
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Answer: D. capillary action
Low surface tension liquid wets the crack and is drawn into it by capillary forces.
161. Dye penetrant testing can detect only:
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The liquid has to get in.
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Answer: A. defects open to the surface
The penetrant must be able to enter the discontinuity, so it works only for surface-breaking flaws.
162. Dye penetrant testing is NOT suitable for:
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Consider what happens on a sponge-like surface.
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Answer: A. very porous or rough surfaces that give a high background indication
Porous surfaces absorb penetrant everywhere, masking real indications; the method itself works on non-porous metals and non-metals.
163. Magnetic particle testing can be applied only to:
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The part must be able to carry magnetic flux.
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Answer: B. ferromagnetic materials such as carbon steel and iron
The method relies on the magnetic flux leakage at discontinuities, which requires a ferromagnetic material.
164. In magnetic particle inspection, a crack is revealed because:
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Particles gather where flux escapes.
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Answer: C. it causes magnetic flux leakage that attracts and holds the iron particles
Flux leaks out at the discontinuity, creating poles that gather the fine particles into a visible indication.
165. For best detection in magnetic particle testing, the magnetic field direction should be:
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Cross the crack at a right angle.
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Answer: A. perpendicular to the length of the expected crack
A crack perpendicular to the field interrupts the flux the most, giving the strongest leakage.
166. A long central conductor carries a current of 1000 A to circularly magnetise a part. At 50 mm from its axis the field strength H = I/(2πr) is about:
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Use r in metres.
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Answer: B. 3183 A/m
H = 1000/(2π × 0.05) = 3183 A/m.
167. Radiographic (X-ray or gamma ray) testing detects internal flaws because:
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Think of a shadow picture.
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Answer: D. the amount of radiation transmitted varies with the thickness and density of material, producing a shadow image on film or detector
Voids and inclusions absorb different amounts of radiation than sound metal, so darker or lighter regions appear in the image.
168. In a radiograph, a gas pore (cavity) in a weld appears as a:
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Less material means more radiation reaches the film.
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Answer: C. darker spot on the film, since more radiation passes through it
A void absorbs less radiation than the surrounding metal, exposing the film more and making it darker.
169. Radiation intensity falls as I = I0 e^(−μx). For μ = 0.1 per mm, the fraction transmitted through 20 mm of steel is:
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Multiply μ and thickness first.
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Answer: D. 13.5 %
I/I0 = e^(−0.1 × 20) = e^(−2) = 0.135.
170. For a material with linear attenuation coefficient μ = 0.462 per cm, the half-value layer (HVL) is:
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HVL = 0.693/μ.
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Answer: B. 1.5 cm
HVL = ln 2/μ = 0.693/0.462 = 1.5 cm.
171. Geometric unsharpness Ug = F·d/D, where F is focal spot size, d the object-to-film distance and D the source-to-object distance. For F = 3 mm, d = 20 mm and D = 600 mm, Ug is:
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Multiply F and d, then divide by D.
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Answer: A. 0.1 mm
Ug = 3 × 20/600 = 0.1 mm.
172. Which precaution is essential when carrying out industrial radiography?
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Reduce the dose to the operator.
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Answer: B. Limit exposure by distance, time and shielding, and keep personnel outside the controlled area
Ionising radiation is hazardous; the principles of time, distance and shielding restrict dose.
173. Eddy current testing can be applied only to materials that are:
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Currents must be able to flow.
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Answer: B. electrically conductive
Eddy currents are induced in conductors by an alternating magnetic field; non-conducting materials cannot be tested.
174. In eddy current testing, a flaw such as a surface crack is detected because it:
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The coil impedance is monitored.
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Answer: B. alters the flow path of eddy currents and changes the impedance of the test coil
A crack interrupts the eddy currents, which alters the secondary magnetic field and hence the coil impedance.
175. The standard depth of penetration of eddy currents is δ = 1/√(πfμσ). For aluminium (σ = 3.5 × 10⁷ S/m, μ = μ0) at 1 kHz it is about:
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δ falls as frequency, permeability and conductivity increase.
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Answer: A. 2.7 mm
δ = 1/√(π × 1000 × 4π×10⁻⁷ × 3.5×10⁷) = 1/√(1.382×10⁵) = 2.69 × 10⁻³ m = 2.7 mm.
176. To inspect for flaws deeper below the surface with eddy current testing, one should:
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Skin depth is inversely related to frequency.
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Answer: C. use a lower test frequency
The skin depth increases as frequency decreases, allowing deeper penetration but at reduced sensitivity to small surface flaws.